0707MediumArrayStringBit Manipulation

Pairing Ribbons on the Label Printer

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Trains the technique from

LeetCode 318Maximum Product of Word Lengths

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A label printer is loaded with a set of ribbons. Ribbon i carries the code codes[i], written in lowercase English letters.

The printer has one ink channel per letter of the alphabet, so two ribbons can run together only when no letter appears on both of them. The span of such a pairing is the product of the lengths of the two codes.

Return the largest span over all pairings the printer can run. If no two ribbons can run together, return 0.

Examples

Example 1

Input: codes = ["mint", "sable", "orchid", "gum"]
Output: 30

The ribbons carrying `sable` and `orchid` share no letter, so they can run together, and their lengths multiply to `5 * 6 = 30`.

Example 2

Input: codes = ["ember", "brace", "camber"]
Output: 0

`ember` and `brace` both use `b`, `ember` and `camber` both use `m`, and `brace` and `camber` both use `c`. No pairing can run, so the answer is 0.

Example 3

Input: codes = ["ox", "zip"]
Output: 6

The only pairing uses `ox` and `zip`, which share no letter, so the span is `2 * 3 = 6`.

Constraints

2 <= codes.length <= 1000
1 <= codes[i].length <= 1000
codes[i] consists only of lowercase English letters.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def best_ribbon_pairing(codes: list[str]) -> int:

Java

public int bestRibbonPairing(String[] codes)

The intended approach

Time O(L + k^2) for k codes with L characters in total: one pass to fold each code, then one AND per pair.Space O(k) for the per-code letter sets and lengths.

Which letters a code uses is all that matters, and there are only 26 of them, so each code folds into a 26-bit set with one bit per letter present. Two codes can then be tested for a shared letter with a single bitwise AND against zero, which makes the pairwise scan cheap no matter how long the codes are. Keep the largest product of lengths over the pairs that pass, starting from 0 so an unpairable set answers 0 on its own.

Pythonentry point: best_ribbon_pairing

def best_ribbon_pairing(codes):
    # O(L + k^2) for k codes of L characters in total: fold each code into a
    # 26-bit letter set, then test each pair with one bitwise AND.
    k = len(codes)
    masks = [0] * k
    sizes = [0] * k
    for i, code in enumerate(codes):
        bits = 0
        for ch in code:
            bits |= 1 << (ord(ch) - 97)
        masks[i] = bits
        sizes[i] = len(code)
    best = 0
    for i in range(k):
        for j in range(i + 1, k):
            if masks[i] & masks[j] == 0:
                span = sizes[i] * sizes[j]
                if span > best:
                    best = span
    return best

The obvious version

For every pair of ribbons, walk the 26 letters and ask whether each one occurs in both codes by searching the strings directly, then keep the largest product of lengths among the pairs with no letter in common.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_best_ribbon_pairing(codes):
    alphabet = "abcdefghijklmnopqrstuvwxyz"
    k = len(codes)
    best = 0
    for i in range(k):
        for j in range(i + 1, k):
            shared = False
            for ch in alphabet:
                if ch in codes[i] and ch in codes[j]:
                    shared = True
                    break
            if not shared:
                span = len(codes[i]) * len(codes[j])
                if span > best:
                    best = span
    return best

How this problem was checked

Reference solution executed against 17 test cases — 17 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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